If two vertices induce isomorphic subgraphs when they are removed, are they conjugate?

Question

This feels like a very natural question, but I am struggling to find a reference, proof, or counter-example.

Consider a simple graph $G$, and say two vertices are conjugate if there exists an automorphism of $G$ that switches them. It is not hard to show that if $v,w$ are conjugate than the induced subgraphs on $G\setminus v$ and $G\setminus w$ are isomorphic.

But is the converse true? In my context of interest, multiple edges between vertices are allowed and all edges are undirected (although I don't anticipate that this affects the answer).

An equivalent statement is that for any isomorphism $\phi$ from $G\setminus \{v\}$ to $G\setminus \{w\}$ and vertex $u$ in $G\setminus \{v\}$, $\phi(u)$ and $u$ have the same degree in the original graph $G$.

Answer 1

Pick any vertex transitive graph $G$ (of degree at least 2) that has vertices $u,v \in V(G)$ such that no automorphism of $G$ swaps $u$ and $v$ (in other words, $G$ is vertex transitive but not generously transitive). Create $G'$ by adding vertices $u'$ and $v'$ to $G$ and making them adjacent to $u$ and $v$ respectively. Then clearly $G'-u'$ and $G'-v'$ are isomorphic but there is no automorphism that maps $u'$ to $v'$ since its restriction to $V(G)$ would have to be an automorphism of $G$ that swaps $u$ and $v$.

Cayley graphs with connection sets that are unions of conjugacy classes (e.g. Cayley graphs for abelian groups) are generously transitive, but I suspect that if you pick a random one without this property then it will very likely not be generously transitive (of course it will be vertex transitive).